# Class 11 RD Sharma solutions – Chapter 29 Limits – Exercise 29.7 | Set 2

### Question 32. limxâ†’0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]

Solution:

We have,

limxâ†’0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]

=

=

= -2sina Ã— (1/2)

= -sina

### Question 33. limxâ†’0[{x2 – tan2x}/(tanx)]

Solution:

We have,

limxâ†’0[{x2-tan2x}/(tanx)]

Dividing numerator by 2x and denominator by x.

=

=

= 2(0 – 1)/1

= -2

### Question 34. limxâ†’0[{âˆš2 – âˆš(1 + cosx)}/x2]

Solution:

We have,

limxâ†’0[{âˆš2 – âˆš(1 + cosx)}/x2]

On rationalizing numerator

= limxâ†’0[{2-(1+cosx)}/x2{âˆš2+âˆš(1+cosx)}]

= limxâ†’0[(1-cosx))/x2{âˆš2+âˆš(1+cosx)}]

= 2 Ã— (1/4) Ã— [1/{âˆš2 + âˆš(1 + 1)}]

= (2/4) Ã— (1/2âˆš2)

= (1/4âˆš2)

### Question 35. limxâ†’0[{xtanx}/(1 – cosx)]

Solution:

We have,

limxâ†’0[{xtanx}/(1 – cosx)]

On dividing the numerator and denominator by x2

=

=

As we know that limxâ†’0[sinx/x] = 1 and limxâ†’0[tanx/x] = 1

= (4/2)

= 2

### Question 36. limxâ†’0[{x2 + 1 – cosx}/(xsinx)]

Solution:

We have,

limxâ†’0[{x2 + 1 – cosx}/(xsinx)]

= limxâ†’0[{x2 + 2sin2(x/2)}/(xsinx)]

On dividing the numerator and denominator by x2

As we know that limxâ†’0[sinx/x] = 1

= 3/2

### Question 37. limxâ†’0[sin2x{cos3x – cosx}/(x3)]

Solution:

We have,

limxâ†’0[sin2x{cos3x – cosx}/(x3)]

As we know that limxâ†’0[sinx/x] = 1

= -2 Ã— 2 Ã— 2

= -8

### Question 38. limxâ†’0[{2sinxÂ° – sin2xÂ°}/(x3)]

Solution:

We have,

limxâ†’0[{2sinxÂ°-sin2xÂ°}/(x3)]

= limxâ†’0[{2sinxÂ°-2sinxÂ°cosxÂ°}/(x3)]

= limxâ†’0[2sinxÂ°{1-cosxÂ°}/(x3)]

= limxâ†’0[2sinxÂ°{2sin2(xÂ°/2)}/(x3)]

= 4 Ã— [Ï€3/(180 Ã— 360 Ã— 360)]

= (Ï€/180)3

### Question 39. limxâ†’0[{x3.cotx}/(1 – cosx)]

Solution:

We have,

limxâ†’0[{x3.cotx}/(1 – cosx)]

= limxâ†’0[x3/{tanx(1 – cosx)}]

As we know that limxâ†’0[sinx/x] = 1 and limxâ†’0[tanx/x] = 1

= 2

### Question 40. limxâ†’0[{x.tanx}/(1 – cos2x)]

Solution:

We have,

limxâ†’0[{x.tanx}/(1 – cos2x)]

= limxâ†’0[{x.tanx}/(2sin2x)]

On dividing the numerator and denominator by x2

As we know that limxâ†’0[sinx/x] = 1 and limxâ†’0[tanx/x] = 1

= (1/2)

### Question 41. limxâ†’0[{sin(3 + x) – sin(3 – x)}/x]

Solution:

We have,

limxâ†’0[{sin(3 + x) – sin(3 – x)}/x]

=

= 2Limxâ†’0[cos3.sinx/x]

= 2cos Ã— 3limxâ†’0[sinx/x]

As we know that limxâ†’0[sinx/x] = 1

= 2cos3

### Question 42. limxâ†’0[{cos2x – 1)}/(cosx – 1)]

Solution:

We have,

limxâ†’0[{cos2x – 1)}/(cosx – 1)]

= limxâ†’0[(2sin2x)/{2sin2(x/2)}]

= limxâ†’0[(sin2x)/{sin2(x/2)}]

As we know that limxâ†’0[sinx/x] = 1

= (x2) Ã— (4/x2)

= 4

### Question 43. limxâ†’0[{3sin2x – 2sinx2)}/(3x2)]

Solution:

We have,

limxâ†’0[{3sin2x – 2sinx2)}/(3x2)]

= limxâ†’0[(3sin2x/3x2) – (2sinx2/3x2)]

As we know that limxâ†’0[sinx/x] = 1

= 1 – 2/3

= (3 – 2)/3

= (1/3)

### Question 44. limxâ†’0[{âˆš(1 + sinx) – âˆš(1 – sinx)}/x]

Solution:

We have,

limxâ†’0[{âˆš(1 + sinx) – âˆš(1 – sinx)}/x]

On rationalizing numerator.

= limxâ†’0[{(1 + sinx) – (1 – sinx)}/x{âˆš(1 + sinx) + âˆš(1 – sinx)}]

= limxâ†’0[2(sinx)/x{âˆš(1 + sinx) + âˆš(1 – sinx)}]

As we know that limxâ†’0[sinx/x] = 1

= 2 Ã— {1/(âˆš1 + âˆš1)}

= 2/2

= 1

### Question 45. limxâ†’0[(1 – cos4x)/x2]

Solution:

We have,

limxâ†’0[(1 – cos4x)/x2]

= limxâ†’0[2sin22x/x2]

As we know that limxâ†’0[sinx/x] = 1

= 2 Ã— 4

= 8

### Question 46. limxâ†’0[(xcosx + sinx)/(x2 + tanx)]

Solution:

We have,

limxâ†’0[(xcosx + sinx)/(x2 + tanx)]

= limxâ†’0[x(cosx+sinx/x)/x(x + tanx/x)]

= limxâ†’0[(cosx + sinx/x)/(x + tanx/x)]

As we know that limxâ†’0[tanx/x] = 1

= (1 + 1)/(1 + 0)

= 2

### Question 47. limxâ†’0[(1 – cos2x)/(3tan2x)]

Solution:

We have,

limxâ†’0[(1 – cos2x)/(3tan2x)]

= limxâ†’0[2sin2x/3tan2x]

=

= (2/3)limxâ†’0[cos2x]

= (2/3)

### Question 48. limÎ¸â†’0[(1 – cos4Î¸)/(1 – cos6Î¸)]

Solution:

We have,

limÎ¸â†’0[(1 – cos4Î¸)/(1 – cos6Î¸)]

= limÎ¸â†’0[2sin22Î¸/2sin23Î¸]

= limÎ¸â†’0[sin22Î¸/sin23Î¸]

=

= [(4Î¸2)/(9Î¸2)]

= (4/9)

### Question 49. limxâ†’0[(ax + xcosx)/(bsinx)]

Solution:

We have,

limxâ†’0[(ax + xcosx)/(bsinx)]

On dividing the numerator and denominator by x

As we know that limxâ†’0[sinx/x] = 1

=(a + cos 0)/b Ã— 1

= (a + 1)/b

### Question 50. limÎ¸â†’0[(sin4Î¸)/(tan3Î¸)]

Solution:

We have,

limÎ¸â†’0[(sin4Î¸)/(tan3Î¸)]

=

As we know that limxâ†’0[sinx/x] = 1 and limxâ†’0[tanx/x] = 1

= (4Î¸/3Î¸)

= (4/3)

### Question 51. limxâ†’0[{2sinx – sin2x}/(x3)]

Solution:

We have,

limxâ†’0[{2sinx – sin2x}/(x3)]

= limxâ†’0[{2sinx – 2sinxcosx}/(x3)]

= limxâ†’0[2sinx{1 – cosx}/(x3)]

= limxâ†’0[2sinx{2sin2(x/2)}/(x3)]

As we know that limxâ†’0[sinx/x] = 1

= (4/4)

= 1

### Question 52. limxâ†’0[{1 – cos5x}/{1 – cos6x}]

Solution:

We have,

limxâ†’0[{1 – cos5x}/{1 – cos6x}]

=

As we know that limxâ†’0[sinx/x] = 1

= 25/(4 Ã— 9)

= (25/36)

### Question 53. limxâ†’0[(cosecx – cotx)/x]

Solution:

We have,

limxâ†’0[(cosecx – cotx)/x]

= limxâ†’0[(1/sinx – cosx/sinx)/x]

= limxâ†’0[(1 – cosx)/x.sinx]

= limxâ†’0[2sin2(x/2)/x.sinx]

As we know that limxâ†’0[sinx/x] = 1

= 2/4

= 1/2

### Question 54. limxâ†’0[(sin3x + 7x)/(4x + sin2x)]

Solution:

We have,

limxâ†’0[(sin3x + 7x)/(4x + sin2x)]

As we know that limxâ†’0[sinx/x] = 1

= (7 + 3)/(4 + 2)

= 10/6

= 5/3

### Question 55. limxâ†’0[(5x + 4sin3x)/(4sin2x + 7x)]

Solution:

We have,

limxâ†’0[(5x + 4sin3x)/(4sin2x + 7x)]

=

=

=

As we know that limxâ†’0[sinx/x] = 1

= (5 + 4 Ã— 3)/(4 Ã— 2 + 7)

= (17/15)

### Question 56. limxâ†’0[(3sinx – sin3x)/x3]

Solution:

We have,

limxâ†’0[(3sinx – sin3x)/x3]

= limxâ†’0[{3sinx – (3sinx – 4sin3x)/x3]

= limxâ†’0[(4sin3x)/x3]

= 4Limxâ†’0[{(sinx)/x}3]

As we know that limxâ†’0[sinx/x] = 1

= 4 Ã— 1

= 4

### Question 57. limxâ†’0[(tan2x – sin2x)/x3]

Solution:

We have,

limxâ†’0[(tan2x – sin2x)/x3]

= limxâ†’0[(sin2x/cos2x-sin2x)/x3]

= limxâ†’0[(2sin2x.sin2x)/(x3cos2x)]

As we know that limxâ†’0[sinx/x] = 1

= 2 Ã— 2/cos0

= 4

### Question 58. limxâ†’0[(sinax + bx)/(ax + sinbx)]

Solution:

We have,

limxâ†’0[(sinax + bx)/(ax + sinbx)]

As we know that limxâ†’0[sinx/x] = 1

= (1 Ã— a + b)/(a + 1 Ã— b)

= (a + b)/(a + b)

= 1

Question 59. limxâ†’0[cosecx-cotx]

Solution:

We have,

limxâ†’0[cosecx – cotx]

= limxâ†’0[1/sinx – cosx/sinx]

= limxâ†’0[(1 – cosx)/sinx]

= limxâ†’0[{2sin2(x/2)}/{2sin(x/2)cos(x/2)}]

= limxâ†’0[sin(x/2)/cos(x/2)]

= limxâ†’0[tan(x/2)/ x/2] Ã— x/2

As we know that limxâ†’0[tanx/x] = 1

= 0

### Question 60. limxâ†’0[{sin(Î± + Î²)x + sin(Î± – Î²)x + sin2Î±x}/{cos2Î²x – cos2Î±x}]

Solution:

We have,

limxâ†’0[{sin(Î± + Î²)x + sin(Î± – Î²)x + sin2Î±x}/{cos2Î²x – cos2Î±x}]

=

= limxâ†’0[{2sinÎ±x.cosÎ²x + 2sinÎ±x.cosÎ±x}/(sin2Î±x – sin2Î²x)]

= limxâ†’0[{2sinÎ±x(cosÎ²x + cosÎ±x)}/(sin2Î±x – sin2Î²x)]

As we know that limxâ†’0[sinx/x] = 1

= [{2 Ã— Î± Ã— 1 Ã— (1 + 1)}/(Î±2 – Î²2)] Ã— (1/0)

= (1/0)

= âˆž

### Question 61. limxâ†’0[(cosax – cosbx)/(cosecx – 1)]

Solution:

We have,

limxâ†’0[(cosax – cosbx)/(cosecx – 1)]

=

=

=

= [(a + b)(a – b)/c2] Ã— (4/4)

= (a2 – b2)/c2

### Question 62. limhâ†’0[{(a + h)2sin(a + h) – a2sina}/h]

Solution:

We have,

limhâ†’0[{(a + h)2sin(a + h) – a2sina}/h]

= limhâ†’0[{(a+h)2(sina.cosh)+(a+h)2(cosa.sinh)-a2sina}/h]

= limhâ†’0[{(a2+2ah+h2)(sina.cosh)-a2sina+(a+h)2(cosa.sinh)}/h]

= limhâ†’0[{a2sina(cosh-1)+2ah.sina.cosh+h2sina.cosh+(a+h)2cosa.sinh}/h]

= limhâ†’0[{a2sina(-2sin2(h/2))+2ah.sina.cosh+h2sina.cosh+(a+h)2cosa.sinh}/h]

=

= 0 + 2asina + 0 + a2cosa

= 2a + a2cosa

### Question 63. If limxâ†’0[kx.cosecx] = limxâ†’0[x.coseckx], find K.

Solution:

We have,

limxâ†’0[kx.cosecx] = limxâ†’0[x.coseckx]

limxâ†’0[kx/sinx] = limxâ†’0[x/sinkx]

klimxâ†’0[x/sinx] = limxâ†’0[kx/sinkx](1/k)

k = (1/k)

k2 = 1

k = Â±1

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